Question

$12$ different toys are to be distributed to three children equally. In how many ways this can be done?

Answer 1

Since we have to divide $12$ different toys among three children equally, we can try to select three groups of toys that each contain four toys each for each child. Use the concept of combinations to find the number of ways to do that. First, select four from $12$ toys. Then four form the remaining eight toys. Now multiply the results as it is done simultaneously.

Complete step-by-step answer:
Here we have to distribute $12$ different toys among three children equally. So, each of the three children should get $\dfrac{{12}}{3} = 4$ toys. Hence, we have to find the total number of different ways in which this can be.
Let’s first understand the concept of combinations. A combination is a mathematical technique that determines the number of possible arrangements in a collection of items where the order of the selection does not matter. In combinations, you can select the items in any order.
Now suppose we want to choose $k$ objects from $n$ objects, then the number of combinations of $k$ objects chosen from $n$ objects is denoted by {}^n{C_k}{\text{ or }}\left( {\begin{array}{*{20}{c}} n \\\ k \end{array}} \right) , it follows that:
\Rightarrow \left( {\begin{array}{*{20}{c}} n \\\ k \end{array}} \right) = {}^n{C_k} = \dfrac{{n!}}{{k!\left( {n - k} \right)!}}
So, let’s start with selecting $4$ toys to give to the first child. The number of ways to select first $4$ toys from a total of $12$ toys is ${}^{12}{C_4}$
Then for the next child, we need to select another $4$ toys out of the remaining $8$ toys. The number of ways of doing that using combination will be ${}^8{C_4}$ ways.
Now, we have selected $8$ toys from the total of $12$ toys. After that, there are only $4$ toys left, which will be given to the third child. There will be just one way to do this.
As these events are occurring simultaneously. So, for finding the total number of ways of selection we need to multiply these results together.
$\Rightarrow {}^{12}{C_4} \times {}^8{C_4} \times 1 = \dfrac{{12!}}{{4!\left( {12 - 4} \right)!}} \times \dfrac{{8!}}{{4!\left( {8 - 4} \right)!}} \times 1 = \dfrac{{12!}}{{4!8!}} \times \dfrac{{8!}}{{4!4!}} \times 1 = \dfrac{{12!}}{{4!4!4!}}$
Let’s resolve the factorial in the numerator and denominator as:
$\Rightarrow \dfrac{{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10 \times 11 \times 12}}{{1 \times 2 \times 3 \times 4 \times 1 \times 2 \times 3 \times 4 \times 1 \times 2 \times 3 \times 4}} = 5 \times 7 \times 9 \times 10 \times 11 = 34650$
The number of ways by which $12$ different toys are to be distributed to three children equally are $34650$

Note: Remember that all the toys must be considered different and should be distributed according to that. If the order of the selected items mattered, then we had to use permutations, i.e. ${}^n{P_k} = \dfrac{{n!}}{{\left( {n - k} \right)!}}$ . An alternative approach for this problem can be by dividing the $12$ different toys into $3$ different groups. The number of ways for doing that can be written as $\dfrac{{12!}}{{4!4!4!}}$.